Question: The sum of two fractions is $\frac{11}{12}$ and their product is $\frac{1}{6}$. What is the lesser of the two fractions? Express your answer as a common fraction.
Explanation: We can make use of the fact that the sum of the roots of the quadratic equation $ax^2 + bx + c = 0$ is $-b/a$ and the product of the roots is $c/a$.  Choosing $a$, $b$, and $c$ so that $-b/a=11/12$ and $c/a=1/6$, we find that the fractions are the solutions to $12x^2 - 11x + 2=0$. Factoring this, we get \[ 12x^2 - 11x + 2 = (3x - 2)(4x - 1). \] Therefore, the solutions of $12x^2 - 11x + 2=0$ are $x=\frac{1}{4}$ and $x=\frac{2}{3}$.  The smaller of these fractions is $\boxed{\frac{1}{4}}$.

An alternative way to obtain the equation $12x^2 - 11x + 2=0$ is to begin with the given equations $x+y=\frac{11}{12}$ and $xy=\frac{1}{6}$.  Solve the first equation for $y$ and substitute $y=\frac{11}{12}-x$ into the second equation.  Distributing, clearing denominators, and rearranging gives $12x^2 - 11x + 2=0$. Then we proceed as before.